Dual Nature Of Radiation And Matter

(a) X-ray produced by the electron therefore has maximum photon energy (E_x) equal to energy (E) of the electron.

E = eV …(1)

Where magnitude of charge of the electron ‘e’ = 1.6 × 10^-19 C

Accelerating potential is V = 30 × 10³ V 3

Putting in equation (1) we get,

⇒ E = 1.6 × 10^-19 C × 30 × 10³ J

Also, E_x = hν

Where, E_x = maximum energy of the X-ray

And h = plank’s constant = 6.63 × 10^-34 J. sec

ν = max. frequency of X-ray

⇒ v_max = E_max/h

⇒

⇒ v_max= 7.24 × 10¹⁸ Hz

Max. frequency of X-ray is 7.24 × 10¹⁸ Hz

(b) Minimum wave length (λ) of X-ray is given by the following relation.

E_x = hν = hc/λ

∴ λ = hc / E_x

c = speed of light = 3 × 10⁸ m/s

⇒

⇒ λ = 0.041nm

Minimum wave length of X-ray is equal to 0.041nm

(a) maximum energy (K. E_max) of emitted electron is given by the following equation

K. E_max = hν – W_{0 ………} equation no. 1

Where h = plank’s constant = 6.63 × 10^-34 J. sec

ν = max. frequency of X-ray

W₀ = work function (minimum energy required to emit an electron from neutral atom)

W₀ = 2.14 eV

And, ν = 6 × 10¹⁴ Hz

From equation No. 1

⇒

⇒ KE_max = 0.34 eV

Or in Joules,

KE_max = 0.34 × 1.6 × 10^-19 = 0.546 × 10^-19 J

maximum KE of emitted electron is 0.546 × 10^-19 J

(b) Stopping Potential is calculated by following relation

eV₀ = K.E_max

Magnitude of charge of the electron ‘e’ = 1.6 × 10^-19 C

V₀ = Stopping Potential

⇒

⇒ V₀ = 0.34 V

Stopping Potential of emitted electron is 0.34 V

(c) K.E_max = 1/2 mv²_max

m = mass of the electron = 9.1 × 10^-31 kg

v_max = maximum velocity of electron

v²_max = 2 K.E_max / m

v_max = (2 × 0.546 × 10^-19 J / 9.1 × 10^-31)^1/2

= 347 km /sec

Maximum velocity of emitted electron is 347 km / sec.

The maximum kinetic energy of the emitted electron is equal to work done on the electron to stop it to reach to the anode. Therefore

K.E_max = 1/2 mv²_max = eV₀

Magnitude of charge of the electron ‘e’ = 1.6 × 10^-19 C

V₀ = Stopping Potential

K.E_max = maximum kinetic energy of emitted electron

m = mass of the electron = 9.1 × 10^-31 kg

v_max = maximum velocity of electron

K.E_max = eV₀ = 1.6 × 10^-19 × 1.5 = 1.5 Ev

= 2.4 × 10^-19 J

maximum KE of electron is 2.4 × 10^-19 J

(a) Energy of photon E = hv = hc/λ

Where h = plank’s constant = 6.63 × 10^-34 J. sec

ν = max. frequency of X-ray

c = speed of light = 3 × 10⁸ m/s

λ = minimum wavelength of X-ray

E = 6.63 × 10^-34 × 3 × 10⁸ / 632.8 x10^-9

∴ E = 3.14 × 10^-19 J

Now, p = h / λ

p = momentum of photon

p = 6.63 × 10^-34/632.8 x10^-9

∴ p = 1.05 × 10^-27 kg-m/sec

Maximum energy and momentum of the electron are 3.14 × 10^-19 J and 1.05 × 10^-27 kg-m/sec respectively.

(b) No. of photons ‘n’ per sec. is equal to total power radiated divided by number of photons (it is assumed that power of each and every photon is same and all the photons are falling on the target).

‘n’ = P/E

Where P = radiated power of LASSER

n = 9.42 × 10⁶/3.14 × 10^-19

n = 3 × 10¹⁶ photons/sec.

No. of photons falling per second is equal to 3 × 10¹⁶ photons/sec.

(c) mv = p

m = mass of Hydrogen = mass of proton = 1.67 × 10^-27 kg

v = speed of Hydrogen

p = momentum of photon as well as Hydrogen atom

v = p/m

v = 1.05 × 10^-27/1.67 × 10^-27 = 0.63 m/sec.

Required speed of Hydrogen atom is 0.63 m/sec.

Energy of a photon E = hv = hc / λ

Where h = plank’s constant = 6.63 × 10^-34 J. sec

ν = average frequency of sunlight

c = speed of light = 3 × 10⁸ m/s

λ = average wavelength of sunlight

E = 6.63 × 10^-34 × 3 × 10⁸ / 550 × 10^-9 = 3.61 × 10^-19 J

ϕ = energy flux per unit area (sq. meter) per sec. falling on the earth.

n = let no. of photons falling per sec. per sq. meter on earth surface.

ϕ = n × E

n = ϕ/E

n = 1.388 × 10³/3.61 × 10^-19 = 3.85 × 10²¹ photons/m².sec

no. of photons falling on Earth surface per square meter per sec. is 3.85 × 10²¹ photons/m².sec.

Photoelectric equation

eV₀ = hv - ϕ_{0 …………} eq.01

Where e = Charge on an electron = 1.6 × 10^-19 C

V₀ = Stopping Potential

h = plank’s constant

v = frequency of photon

ϕ₀ = Work function of the metal

Rewriting equation no. 1

V₀ = hv/e - ϕ₀/e ………………………………equ.02

h/e is slop of the line represented by above equation02.

Therefore h/e = slop of the line = 4.12 × 10^-15 Vs

h = 1.6 × 10^-19 × 4.12 × 10^-15 = 6.594 × 10^-34 Js

Value of plank’s constant ‘h’ from above experiment is equal to 6.594 × 10^-34 Js.

(a) Energy of photon E = hv = hc/λ

Where h = plank’s constant = 6.63 × 10^-34 J. sec

ν = frequency of light

c = speed of light = 3 × 10⁸ m/s

λ = wavelength of light

E = 6.63 × 10^-34 × 3 x10⁸ / 589 × 10^-9

= 3.38 × 10^-19 J

Energy of a photon is equal to 3.38 × 10^-19 J.

(b) Rate of photon falling on the sphere = power of lamp/energy per photon

∴ Rate of photon = 100 / 3.38 × 10^-19J = 3.0 × 10²⁰ photons/sec.

Rate at which photons are falling on the sphere is equal to 3.0 × 10²⁰ photons/sec.

Photoelectric equation

eV₀ = hv - ϕ_{0 ………………} equation(1)

Where e = Charge on an electron = 1.6 × 10^-19 C

V₀ = Stopping Potential

h = plank’s constant

v = frequency of photon

ϕ₀ = Work function of the metal

Threshold frequency is the minimum frequency at which photoemission starts taking place from a metal surface. It is represented by v_0.

At threshold frequency, stopping potential is zero and energy of the emitted electron is also zero. Energy provided by the incident photon is just sufficient to eject out electrons from the metal surface.

ϕ₀ = hv₀ ....................equ.02

V₀ = Threshold frequency

Putting value of ϕ₀ from equ.02 to equ.01

eV₀ = hv – hv₀

∴ V₀ = h (v-v₀)/ e = 6.63 × 10^-34 × (8.2-3.3) × 10¹⁴/1.6 × 10^-19

= 2.03 V

For photo emission cut of potential is equal to 2.03 V

For photoemission energy ‘E’ of incident radiation must be more than or equal to work function ‘ϕ₀’ of the metal.

E ϕ₀ (The Condition for photoelectric emission)

E = hv = hc/λ

Where h = plank’s constant = 6.63 × 10^-34 J. sec

ν = frequency of radiation

c = speed of light = 3 × 10⁸ m/s

λ = wavelength of radiation

E = 6.63 × 10^-34 × 3 × 10⁸ / 330 × 10^-9

= 6.018 × 10^-19 J = 6.018 × 10^-19/1.6 × 10^-19 eV

= 3.76 eV

ϕ₀ = 4.2 eV as per question.

E < ϕ_0, therefore photoemission will not take place.

No photoemission will take place as work function is more than the photon energy of the incident radiation.

Photoelectric equation

K. E_max = hv – hv_{0 ……………………………………….}equ.01

Where K.E_max = maximum kinetic energy of the photoelectron

h = plank’s constant

v = frequency of photon

v₀ = threshold frequency

K. E_max = 1/2 mv²_max ........................................equ.02

Where ‘m’ = mass of the electron = 9.1 × 10^-31 kg

V_max = maximum speed of photoelectrons

From equ.01 and equ.02

1/2 mv²_max = hv - hv₀

∴ v₀ = v - 1/2 mv²_max / h

⇒

⇒ v₀ = 7.21 × 10¹⁴ –2.47 × 10¹⁴

⇒ v₀ = 4.74 × 10¹⁴ Hz

Threshold frequency for photoemission is equal to 4.74 × 10¹⁴ Hz.

In photoelectric effect an electromagnetic radiation (Can be Visible Light) is incident on a metal, if the energy of incident radiation is more than work function of the metal, then electrons are ejected out of the metal surface with some kinetic energy

Work function is the minimum Energy required to just pull out an electron from metal surface.

the setup for photoelectric effect has been shown in figure

Now the energy supplied to electron through photon (assuming light to be particle) is used up in overcoming the internal atomic forces to come to surface of metal (equal to work function of metal) and end extra energy is converted to kinetic energy of particle

So the maximum kinetic energy of an electron can be given by the photoelectric equation

K_max = E – ϕ_o

Where K_max is the maximum kinetic energy of electron.

E is the energy of incident radiation.

ϕ_o is the work function of metal.

We know energy of an electromagnetic radiation can be given by relation

E = h𝜈

Where h is Planck’s constant.

𝜈 is the frequency of radiation.

So, above equation becomes

K_max = h𝜈 – ϕ_o

Now to stop the moving electrons a potential difference is such applied which applies force in opposite direction to the motion, if the energy transferred to electron due to potential difference in equal to maximum kinetic energy of electron then the electrons completely stop moving. We know energy transferred or work done by a potential difference on a charged particle is given by relation

W = qV_o

Where W is the energy transferred by a potential difference of V_o volts to a charge particle having charge q

Now for electrons to completely stop we have,

K_max = W

Or K_max = qV_o

Since particle is electron so we have

q = e

e is charge of electron, e = 1.6 × 10^-19 C

i.e. K_max = eV_o

So, the photoelectric equation becomes

eV_o = h𝜈 – ϕ_o

or we can say the work function is given as

ϕ_o = h𝜈 – eV_o

where V_o is stopping Potential Difference

we know the relation for any electromagnetic wave we have

c = 𝜈𝜆

Where 𝜈 is the frequency and 𝜆 is wavelength of electromagnetic radiation having velocity C

So, re arranging we get

𝜈 = c/𝜆

So the photo electric equation becomes

ϕ_o = hc/𝜆 – eV

We know value of Planck’s constant

h = 6.6 × 10^-34Js

charge of electron

e = 1.6 × 10^-19 C

the speed of light

C = 3 × 10⁸ m/s

light is incident from argon laser Light of wavelength

𝜆 = 488 nm = 488 × 10^-9 m

We are given the stopping potential difference

V_o = 0.38 V

Putting thse values in equation to find work function of emitter ϕ_o

ϕ_o = hC/𝜆 – eV_o

we have

We get

ϕ_o = 4.05 × 10^-19J – 0.608 × 10^-19 J = 3.46 × 10^-19J

so work function of emitter metal is 3.46 × 10^-19J

converting it to eV

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get work function of emitter metal as

We get ϕ_o = 2.16 eV

So the work function of emitter metal is 2.16 eV

(a) When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by

K = Vq, which imparts it with velocity

We know kinetic energy of a Particle with mass m kg and velocity v is given by

Equating both equations

We get velocity of the particle as

But we know momentum is given by relation

P = mv

Where P is the momentum of particle having mass m and moving with velocity v

So we get

Here Potential difference

V = 56 volts

q = 1.6 × 10^-19 C

m = 9.1 × 10^-31Kg

so momentum of the electron,

So momentum of electron is 4.04 × 10^-24 Kgms^-1

(b) Now we know de Broglie wavelength of a Particle is given by relation

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v

But we know momentum is given by relation

P = mv

Where P is the momentum of particle having mass m and moving with velocity v

So substituting we get de Broglie wavelength of Particle as

𝜆 = h/P

h = 6.6 × 10^-34Js

p = 4.04 × 10^-24 Kgms^-1

So putting the values in equation we get

So the de Broglie wavelength of electron is 0.164 nm

Note:we could have also used the direct relation when an electron is accelerated through a potential difference of V volts its de Broglie wavelength is given by

1A^o = 10^-10m

Here we are given kinetic energy of electron

K = 120 eV

We know 1 eV = 1.6 × 10^-19 J

i.e. kinetic Energy ,K = 120 × 1.6 × 10^-19 J = 1.92 × 10^-17J

(a) We know kinetic energy of a particle is given by the relation

Where K is the kinetic energy of a particle of mass m moving with speed v

Now multiplying L.H.S. and R.H.S. of the equation by m we get

or

But we know momentum is given by relation, p = mv

So, substituting we get

or

i.e. p = √2mK

Or we can say momentum P of any particle can be expressed in terms of its mass m and kinetic energy K as

p = √2mK

Particle is electron so we have mass of electron

m = 9.1 × 10^-31Kg

kinetic energy of particle

K = 1.92 × 10^-17J

Putting the values in equation we get

= 5.91 × 10^-24 Kgms^-1

So we get momentum of electron is 5.91 × 10^-24 Kgms^-1

(b) But we know momentum is given by relation

P = mv

Where P is momentum of particle of mass m moving with speed v

So we get speed of electron is

v = P/m

here the momentum of electron

P = 5.91 × 10^-24 Kgms^-1

Mass of electron

m = 9.1 × 10^-31Kg

so putting these values we get the speed of electron as

or we can say speed of electron is 6.5 × 10⁶ ms^-1

(c) Now we know de Broglie wavelength of a Particle is given by relation

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v and h is Planck’s constant

But we know momentum is given by relation

P = mv

Where P is the momentum of particle having mass m and moving with velocity v

So substituting we get de Broglie wavelength of Particle as

𝜆 = h/P

value of Planck’s constant

h = 6.6 × 10^-34Js

and momentum of electron is

P = 5.91 × 10^-24 Kgms^-1

Putting these values we get

= 0.112 nm

So the wavelength of the electron is 0.112 nm

Now we know de Broglie wavelength of a Particle is given by relation

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v

But we know momentum is given by relation

p = mv

So, substituting we get de Broglie wavelength of Particle as

𝜆 = h/p

Or we can say momentum of particle is

p = h/ 𝜆

We know kinetic energy of a particle is given by the relation

Where K is the kinetic energy of a particle of mass m moving with velocity v

Now multiplying L.H.S. and R.H.S. of the equation by m we get

or

But we know momentum is given by relation

P = mv

So substituting we get

or

we get kinetic energy of particle K in terms of momentum P and mass of particle m

so putting P = h/ 𝜆

we get kinetic energy of particle as

Where h is the Planck’s constant, m is mass of particle having de Broglie wavelength 𝜆

(a) Now we are given an electron whose de Broglie Wavelength should be

𝜆 = 589 nm = 589 × 10^-9 m

We know mass of electron is, m = 9.1 × 10^-31 Kg

and value of Planck’s constant, h = 6.6 × 10^-34Js

putting values in the relation to find the kinetic energy of electron K we get

So kinetic energy of the electron is 6.9 × 10^-25J

Converting it to eV

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get kinetic energy of electron as

Or we get Kinetic energy of electron as 4.31𝜇eV

(b) Now we are given an neutron whose de Broglie Wavelength should be

𝜆 = 589 nm = 589 × 10^-9 m

We know mass of electron is

m = 1.66 × 10^-27 Kg

and value of Planck’s constant

h = 6.6 × 10^-34Js

putting values in the relation to find the kinetic energy of neutron K we get

So kinetic energy of the electron is 3.78 × 10^-28J

Converting it to eV

We know

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get kinetic energy of electron as

Or we get Kinetic energy of electron as 2.36 neV

Now, we know de Broglie wavelength of Particle can be given by relation

Where, 𝜆 is de Broglie wavelength of a Particle
m, mass of the body
v, velocity of the body
h, planck constant = 6.63 × 10^-34 Js

(a)
given,

mass of bullet, m = 0.040 kg

the speed of bullet is, v = 1.0 km/s = 1000 m/s

putting the valued of m, v, h in the relation

we get the de Broglie wavelength of Bullet as

So, de Broglie wavelength of Bullet is 1.65 × 10^-35 m

(b)
given,
mass of Ball m = 0.060 kg

the speed of Ball is, v = 1.0 m/s

putting the valued of m , v, h in the relation

we get the de Broglie wavelength of Ball as

So, de Broglie wavelength of Ball is 1.1 × 10^-32 m

(c)
given,

mass of dust particle m = 1 × 10^-9 kg

the speed of dust particle is

v = 2.2 m/s

putting the valued of m, v, h

we get the de Broglie wavelength of Ball as

So, de Broglie wavelength of Dust Particle is 3.0 × 10^-25 m

(a) In order to find momentum, we know de Broglie wavelength for any particle is given by

𝜆 = h/P

Or P = h/𝜆

Where h is Planck’s constant

h = 6.63 × 10^-34 Js

and P is the momentum of particle

here we are given the de Broglie wavelength

𝜆 = 1.00 nm = 1.00 × 10^-9 m

Note:we can see that momentum only depends on de Broglie wavelength and Planck’s Constant which is same for both electron or Photon(Light Particle) so the momentum for both will also be same

Putting values we get momentum

So momentum of electron and photon is 6.63 × 10^-25Kgms^-1

(b) we know energy of an photon is given by the relation

E = hc/𝜆

Where E is the energy of Photon having wavelength 𝜆 and speed c,h is Planck’s constant

h = 6.63 × 10^-34 Js

Speed of light or photon in free space

c = 3 × 10⁸ m/s

And we are given wavelength

𝜆 = 1.00 nm = 1.00 × 10^-9 m

Putting these values we get

Converting it to eV

We know

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get energy of Photon as

so we get energy of Photon as 1.24KeV

(c) We know kinetic energy of a particle is given by the relation

Where K is the kinetic energy of a particle of mass m moving with velocity v

Now multiplying L.H.S. and R.H.S. of the equation by m we get

or

But we know momentum is given by relation

P = mv

Where m is mass of Particle moving with velocity v

So substituting we get

or we get kinetic energy of particle K in terms of momentum P and mass of particle m as

The particle is electron and mass of electron is

m = 9.1 × 10^-31Kg

Here the momentum of electron

P = 6.63 × 10^-25Kgms^-1

Putting these values in above equation we have

Converting it to eV

We know

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get Kinetic energy of electron as

so we get Kinetic energy of electron as 1.51eV

Now we know de Broglie wavelength of a Particle is given by relation

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of the particle having mass m and moving with velocity v

But we know momentum is given by relation

P = mv

Where P is the momentum of particle having mass m and moving with velocity v

So substituting we get de Broglie wavelength of Particle as

𝜆 = h/P

Or we can say momentum of particle is

P = h/ 𝜆

We know kinetic energy of a particle is given by the relation

Where K is the kinetic energy of a particle of mass m moving with velocity v

Now multiplying L.H.S. and R.H.S. of the equation by m we get

or

But we know momentum is given by relation

P = mv

So substituting we get

or

we get kinetic energy of particle K in terms of momentum P and mass of particle m

so putting P = h/ 𝜆

we get kinetic energy of particle as

Where h is the Planck’s constant, m is mass of particle having de Broglie wavelength 𝜆

Here,

The de Broglie Wavelength is, 𝜆 = 1.40 × 10^–10m

The particle is neutron and mass of neutron is, m = 1.66 × 10^–27Kg

putting these values we get the kinetic energy of neutron as

So kinetic energy of the neutron is 6.75 × 10^-21J

Converting it to eV

We know

1 eV = 1.6 × 10^-19J

i.e.

1J = (1/1.6 × 10^-19)eV

So we get kinetic energy of neutron as

Or we get Kinetic energy of neutron as 0.042 Ev

Here the neutron is in thermal equilibrium with matter and it have a kinetic due which depends upon temperature of surrounding

We have a relation to find kinetic energy of a atomic particle at any Temperature which is

Where E_K is the kinetic energy of particle at temperature T and K is Boltzmann constant

K = 1.38 × 10^-23 JK^-1

Here temperature is

T = 300 K

So linetic energy of neutron is

So kinetic energy of neutron is

E_k = 6.21 × 10^-21J

And we know Kinetic energy of a particle is given by the relation

Where vis the speed of the particle and m is the mass of the particle

Multiplying both side sides by m

Or

So we get

We know de-Broglie wavelength of a particle is given by relation

where 𝜆 is the de-Broglie wavelength of a particle having mass m and moving with velocity v, h is Planck’s constant

putting the value of

in above equation we get de Broglie wavelength

Here the particle is neutron and we know mass of neutron

m = 1.66 × 10^-27Kg

Value of Planck’s constant

h = 6.63 × 10^-34Js

and kinetic energy of neutron is

E_k = 6.21 × 10^-21J

Putting values in above equation we get

Or 𝜆 = 0.146 × 10^-9m = 0.146 nm

So the de Broglie wavelength of neutron is 0.146 nm

we know the relation between energy and momentum P of a electromagnetic radiation having velocity C and energy E is

P = E/C

and we know energy of electromagnetic radiation is given by relation

E = h𝜈

Where E is the energy of Photon, 𝜈 is the frequency of electromagnetic radiation and h is Planck’s constant

So putting value of E in equation P = E/c

we get

P = h𝜈/c

We know relation between frequency wavelength and velocity of electromagnetic radiation as

c = 𝜈𝜆

Where 𝜈 is the frequency and 𝜆 is wavelength of electromagnetic radiation having velocity c

So re arranging we get

𝜈 = c/𝜆

Putting the value of 𝜈 in equation P = h𝜈/c

We get

P = hc/λc = h/𝜆

Or we can say the wavelength of electromagnetic radiation is given as

𝜆 = h/P

Where h is Planck’s constant and P is momentum of electromagnetic radiation

Now we know de Broglie wavelength of quantum (photon) can be given by relation

𝜆 = h/mv

Where 𝜆 is de Broglie wavelength of a photon having mass m and moving with velocity v (here the velocity of photon v = C)

But we know momentum is given by relation

P = mv.

So, substituting we get de Broglie wavelength of quantum (photon) as

𝜆 = h/P

This is same as wavelength of electromagnetic radiation

We have a relation to find kinetic energy of a gas molecule at any Temperature which is

Where E_K is the kinetic energy of a particle at temperature T and K is boltzmann constant

K = 1.38 × 10^-23 JK^-1

And we know Kinetic energy of a particle is given by the relation

Where v_rms is the root meansquare speed of the particle and m is the mass of the particle

So equating both equations we have

On solving we get the root meansquare speed of the particle as

Where V_rms is the root mean square speed of the particle of mass m at temperature T and K is boltzmann constant

K = 1.38 × 10^-23 JK^-1

Here the temperature is

T = 300K

We are given the particle is a nitrogen molecule (N₂) mass of the Nitrogen atom is

m_N = 14.0076 u

so mass of N₂ molecule is

m = 2 × 14.0076 u = 28.0152 u

converting it to Kg

we know 1 u = 1.66 × 10^-27 Kg

so mass of Nitrogen molecule is

m = 28.0152 × 1.66 × 10^-27Kg = 46.50 × 10^-27Kg

= 4.650 × 10^-26Kg

We know de-Broglie wavelength of a particle is given by relation

where 𝜆 is the de de-Broglie wavelength of a particle having mass m and moving with velocity v , here velocity is the root mean square speed of the particle, h is Planck’s constant

h = 6.63 × 10^-34 Js

putting the value root mean square speed

in above equation we get

Solving we get the de Broglie wavelength is

Putting value of m,T,K and h in above equation we get

Or 𝜆 = 0.028 × 10^-9m = 0.028 nm

So, the de Broglie wavelength of Nitrogen Molecule is 0.028 nm

Given:

Potential difference between collector and emitter = 500V

Specific charge of electron (charge per unit mass e/m) = 1.76 × 10¹¹ C

Kinetic energy of an electron is given by:

…(1)

Where,

M = mass of electron

v = velocity of electron

e = charge of electron

V = potential difference (accelerating potential)

(a) From equation (1), we can write

…(2)

By putting the values in equation (2) we can find electron velocity.

v = 1.327 × 10⁷ ms^-1

(b) Accelerating potential, V = 10MV = 10⁶V

Let speed of electron be v₁

Again putting the values in equation (2),

v₁ =

v_{1 =} 1.8 × 10⁹ms^-1

This result is wrong as we understand that speed of light

(i.e. 3 × 10⁸ ms^-1) is the theoretical limit of the speed.

Such problems can be dealt using relativistic mechanics,

Relativistic mass is given by:

m =

Where,

m = relativistic mass

m₀ = rest mass

v = velocity of particle

c = speed of light

At relativistic speeds, kinetic energy is given by,

KE = mc²-m₀c²

Given:

Speed of electron, v = 5.20 × 10⁶ ms^-1

Magnetic field strength normal to the beam = 1.30 × 10^–4 T

Charge to mass ratio (e/m) = 1.76 × 10¹¹ C Kg^-1

(a) Force applied by magnetic field,

F = e|v × B|

F = e × v × B × sinθ …(1)

Where,

e = Charge on electron

v = velocity of particle

B = magnetic field strength

θ = angle between Magnetic field and velocity

Since the electron traces a circular path, we can use the following equation of centrifugal force:

F = …(2)

Where,

m = mass of particle

v = velocity of particle

r = radius of circle traced

By equating (1) and (2), we get,

evB × sin90° =

r =

→ r = …(3)

→ r =

→ r = 22.7 6 cm

(b) Energy of the electron beam, E = 20 MeV

E = 20 × 10⁶ × 1.6 × 10^-19J

The energy of an electron is given by:

E =

From the above equation, we can write,

v = …(4)

putting the value in equation (4)

v =

v = 2.6 × 10⁹ ms^-1

This result is incorrect because the speed of any massive object can’t exceed the speed of light (i.e. 3 × 10⁸ ms^-1). We can’t use equation (4) in the case where the speed is relativistic:

At relativistic speeds, mass is given by,

m =

Where,

m = relativistic mass

m₀ = rest mass

v = velocity of particle

c = speed of light

the radius of the path traced is given by,

r = …(3)

By substituting the value of relativistic mass in equation (3) we get,

r = …(4)

By using equation (4) we can find the radius traced by electrons moving at relativistic speed.

By substituting value of relativistic mass in equation

Given:

Potential at collector, V = 100V

Magnetic field strength, B = 2.83 × 10^-4 T

Radius of orbit traced by electron, r = 12.0 cm

Let specific charge (i.e. e/m) = K

Kinetic energy of an electron is given by:

KE = = eV

→ v = …(1)

Where,

M = mass of electron

v = velocity of electron

e = charge of electron

V = potential difference (accelerating potential)

We also know that the bending is caused by the magnetic field,

So we can write,

Magnetic force = centrifugal force

e × v × B =

→ e × B =

→ v = …(2)

Where,

e = Charge on electron

v = velocity of particle

B = magnetic field strength

m = mass of electron

r = radius of trajectory

By equating (1) and (2) we can write,

.

→

By putting the values in above equation we get,

Hence the specific charge is 1.7 × 10^-11 C Kg^-1.

Given:

The wavelength of radiation produced by tube, λ = 0.45 Å

i.e. 0.45 × 10^-10 m

(a) The maximum energy of the photon is given by,

…(1)

Where,

E = Energy of photon

h = Planck’s constant = 6.6 × 10^-34 Js

c = 3 × 10⁸ ms^-1

λ = wavelength of photon

by putting the values in equation (1), we get,

E = 44 × 10^-16 J or 27.6 × 10³ eV

The maximum energy of the photon in x-ray is 44 × 10^-16J.

(b) For a photon to have 27 KeV of energy, the accelerating potential must be of the order of 30 KeV.

Given:

Total energy of system, E = 10.2 BeV = 10.2 × 10⁹ × 1.6 × 10^-19J

E = 16.32 × 10^-10J

Energy in each ray, E’ = E/2

E’ = 8.16 × 10^-10J

We know that Energy in a photon is given by,

…(1)

Where,

E = Energy of photon

h = Planck’s constant = 6.6 × 10^-34 Js

c = 3 × 10⁸ ms^-1

λ = wavelength of photon

From equation (1) we can get,

λ = 2.34 × 10^-16m

Hence, the wavelength of each ray is 2.3 × 10^-16 m.

Given:

Power of transmitter, E’ = 10 KW = 10000 Js^-1

Wavelength of radio waves emitted = 500 m

(a) We know that Energy in a wave is given by,

…(1)

Where,

E = Energy of photon

h = Planck’s constant = 6.6 × 10^-34 Js

c = 3 × 10⁸ ms^-1

λ = wavelength

E = 3.9 × 10^-28 J

Total power = number of photons emitted × Energy of photon

E’ = n × E

n =

n = 2.55 × 10³¹ s^-1

n = 3 × 10^-31 s^-1

Here we see that the number of radio waves emitted per second is very high.

(b) Given:

Intensity of light perceived by human eyes, I = 10^-10Wm^-2

Area of pupil, A = 0.4 × 10^-4 m²

Frequency of white light, v = 6 × 10¹⁴Hz

The energy of each photon is given by,

E = hv

Where,

E = energy of photon

h = Planck’s constant = 6.6 × 10^-34Js

v = 6 × 10¹⁴Hz

E = 6.6 × 10^-34Js × 6 × 10¹⁴ s^-1

E = 3.96 × 10^-19J

Energy of each photon is 3.96 × 10^-19J

Let total number of photons being emitted per second, falling on unit area = n

We define intensity as the amount of energy falling on unit area in unit time so we can write,

I = n × 3.96 × 10^-19J

10^-10 Jm^-2s^-1 = n × 3.96 × 10^-19J

n = 2.52 × 10⁸ m^-2s^-1

Number of photons entering pupil = area of pupil × n

Number of photons entering pupils = 0.4 × 10^-4m² × 2.52 × 10⁸ m^-2s^-1

→ n_p = 1.0082 × 10⁴s^-1

Almost 10000 photons enter our pupil per second.

Given:

Wavelength of light, λ = 2271 Å = 2271 × 10^-10m

Power of mercury source, E = 100 Js^-1

Stopping potential, V_s = -1.3 V

Let frequency of light = v

Work function, Φ₀ is given by,

Φ₀ = hv-eV_s

Φ₀ =

Where,

h = Planck’s constant = 6.6 × 10^-34Js

c = speed of light = 3 × 10⁸m

λ = wavelength of light

e = charge on each electron = 1.6 × 10^-19C

Φ₀ =

Φ₀ = 6.64 × 10^-19J

Φ₀ = (6.64/1.6) × 10^-19 eV = 4.15 eV

Let v₀ be the threshold frequency of the metal,

Φ₀ = hv₀

v_{0 =} Φ₀/h

→ v₀ = 6.6 × 10^-19/ 6.6 × 10^-34

→ v₀ = 1.00. × 10^-15 s^-1

Wavelength of red light, λ’ = 6323 × 10^-10m

Frequency of red light can be given as,

v' = c/λ’

v’ =

v’ = 4.74 × 10¹⁴ Hz

Since the threshold frequency is greater than the frequency of red light, the photocell will not respond to the red light produced.

Given:

Wavelength of light, λ = 640.2 nm = 640.2 × 10^-9m

Stopping potential, V₀ = 0.54 V

Let frequency of light = v

Work function, Φ₀ is given by,

Φ₀ = hv-eV_s

Φ₀ = …(1)

Where,

h = Planck’s constant = 6.6 × 10^-34Js

c = speed of light = 3 × 10⁸m

λ = wavelength of light

e = charge on each electron = 1.6 × 10^-19C

Φ₀ =

Φ₀ = 2.229 × 10^-19J or 1.39 eV

Wavelength of radiation emitted by iron, λ’ = 427.2 nm

λ’ = 4.27.2 × 10^-9m

The new stopping potential can be found using equation (1),

Let new stopping potential = v’

By putting the values in equation (1) we get,

Φ₀ =

→

→

→ v’ = 1.5 eV

Therefore, the new stopping potential is 1.5 eV.

Given:

The frequency, v of a wave with wavelength λ is given by,

v = c/λ …(1)

(c = speed of light = 3 × 10⁸ms^-1)

By using equation (1) we can find frequency for each case,

The equation for photo-electric effect is given as,

Φ₀ = hv- eV₀

We can rewrite this equation as,

V₀ = …(2)

Where,

h = Planck’s constant

c = speed of light = 3 × 10⁸m

λ = wavelength of light

e = charge on each electron = 1.6 × 10^-19C

We find that the slope of the graph remains same,

Slope of the line =

From equation (2),

We can write,

Slope = h/e

h = e × slope

h =

h = 6.58 × 10^-34 Js

From the same graph, we can infer that, threshold frequency of the metal is 5 × 10¹⁴ Hz

i.e. v₀ = 5 × 10¹⁴Hz

Work function, Φ₀ = hv₀ …(2)

Where,

h = Plank’s constant = 6.57 × 10^-34Js

v₀ = threshold frequency

By plugging the data in equation (2), we get,

Φ₀ = 6.57 × 10^-34Js × 5 × 10¹⁴Hz

→ Φ₀ = 3.28 × 10^-19 J

→

→ Φ₀ = 2eV

Hence, the work function of the metal is 2eV.

Given:

Wavelength of radiation from laser = 3300 Å = 3300 × 10^-10m

Energy radiated by the laser, E is given by,

…(1)

Where,

E = Energy of photon

h = Planck’s constant = 6.6 × 10^-34 Js

c = 3 × 10⁸ ms^-1

λ = wavelength of photon

putting values in equation (1) we get,

E = 6 × 10^-19 J

E = (6/1.6) × 10^-19

E = 3.158 eV

We observe that Na and K have work functions less than the energy radiated by the laser so photoelectric effect will take place in case of them, whereas no photo-electric effect will occur in case of Mo and Ni.

Even if the laser is brought closer, it will have no effect on the outcome as the photo-electric effect is independent of the intensity of light.

Given:

Intensity of light, I = 10^-5 Wm^-2

Surface area of photo-cell, A = 2 cm² = 0.0002 m²

Work function of sodium, Φ₀ = 2eV = 2 × 1.6 × 10^-19 = 3.2 × 10^-19J

By knowing the effective area of each sodium atom we can find the absorption of incident energy,

Effective area of each sodium atom, A_Na = 10^-20 m²

Number of layers of sodium, n = 5

Number of atoms absorbing radiation, N_Na = n × (A_Na/ A)

→ N_Na = 5 × (10^-20 m²/ 0.0002 m²)

→ N_Na = 10¹⁷

So we conclude that 10¹⁷ atoms are effectively absorbing radiation.

Energy absorbed per atom, E = I/N_Na

E = 10^-5Js^-1/10¹⁷

→ E = 2 × 10^-26 Js^-1

Times required for photo electric emission, t is give by,

t = Φ₀/E

t = 3.2 × 10^-19 J / 2 × 10^-26 Js^-1

t = 1.6 × 10⁷ s

t = 0.5 years

Hence the time requires to initiate photoelectric emission is 0.5 years which is impractical hence, the wave model stands in disagreement with the experimental results.

Given:

Wavelength of probe, λ = 1 Å = 10^-10m

Mass of electron, m_e = 9.11 × 10^-31Kg

The kinetic energy of an electron is given by:

We can write,

mv = (2 × m × KE)^0.5

P = (2 × m × KE)^0.5 ...(1)

Where,

m = mass of electron

v = velocity of electron

p = momentum of particle

De-broglie wavelength is given by,

Where,

λ = Wavelength

h = Planck’s constant

p = momentum

From equation (1) we can write,

…(2)

From equation (2) we can write that,

KE =

Putting the values in above equation we get,

KE =

KE = 2.39 × 10^-17 J

KE for an electron is 2.39 × 10^-17 J.

Now for photon,

E’ = …(4)

Where,

E = energy of photon

h = Planck’s constant

c = speed of light

λ = wavelength of the photon

By putting values in equation (4 ) we get,

E’ = 19.6 × 10^-16 J

The photon will have more energy as compared to the accelerated electron for the same wavelength.

Given:

Kinetic energy of neutron, E = 150 eV = 150 × 1.6 × 10^-19

E = 2.4 × 10^-17 J

Mass of neutron, m_n = 1.675 × 10^-27 Kg

Kinetic energy of a particle is given by,

We can write,

mv = (2 × m × KE)^0.5

P = (2 × m × KE)^0.5 ...(1)

Where,

m = mass of electron

v = velocity of electron

p = momentum of particle

By putting the equation (1) in de- Broglie equation, we get,

…(2)

Now substituting the values in eq (2) we get,

λ = 2.327 × 10^-12m

A neutron cannot be used in diffraction experiment as lattice spaces are of the order of few 10^-10 m, whereas the wavelength of a neutron beam is 100 times smaller.

De- Broglie wavelength, λ = 1.447 × 10^-10 m

Room temperature, T = 300K

The average KE of neutron is given as,

KE = 3/2 KT …(3)

Where,

K = Boltzmann constant = 1.38 × 10^-23 Jmol^-1K^-1

From equation 2,

Using equation (3) we can write,

By putting values in above equation we get,

λ = 1.447 × 10^-10 m

Now, the wavelength has come into comparable range with the inter-atomic spaces, it is now suitable for diffraction. Hence a beam should be thermalized before using for diffraction.

Given:

Accelerating voltage, V = 50 KV

Energy of each electron is given by,

E = eV …(1)

Where,

e = Charge on electron = 1.6 × 10^-19C

V = accelerating Voltage

Putting the values in equation (1)

E = 1.6 × 10^-19C × 50 × 1000V

E = 8 × 10^-15 J

De Broglie wave length is given by,

Where,

m = mass of electron

h = Planck’s constant = 6.6 × 10^-19 Js

E = Energy

By putting the values in above equation we get,

λ = 5.4 × 10^-12 m.

As compared to an optical telescope which uses yellow light, electron microscope has better resolving power (about 10⁵ times better). As the resolving power is inversely proportional to the wavelength of light used.

Given:

Order of length of quark structure, λ = 10^-15m

Rest mass energy of electron = 0.511 MeV

i.e. m₀c² = 0.511 MeV

→ m₀c² = 0.511 × 1.6 × 10^-19

→ m₀c² = 0.817 × 10^-13 J

By using De-Broglie wavelength equation we can write,

Where,

p = momentum of particle

h = Planck’s constant = 6.6 × 10^-34Js

λ = wavelength

The equation of energy at relativistic speed is,

E² = p²c² + m²₀c⁴

E² = (6.6 × 10^-19 × 3 × 10⁸)² + (0.817 × 10^-13)²

→ E = (392.07 × 10^-22)^0.5

→ E = 1.9 × 10^-10J

Hence the energy of electron emitted from linear accelerator is 1.9 × 10^-10J.

Given:

Pressure, P = 1 atm = 101325 Pa

Temperature, T = 27° = 300K

De-Broglie wavelength is given by,

…(1)

Where,

h = Planck’s constant = 6.6 × 10^-34 Js

m = mass of He

E = total energy

Mass of helium, m = Atomic mass/ number of atoms

Let there be one mole of He,

mass of 1 mol of He = 4 g

Then,

m = 4/ 6.023 × 10²³

m = 6.64 × 10^-24 = 6.64 × 10^-27Kg …(2)

We know that average energy at temperature T is given by,

E = 3/2 kT …(3)

Where,

k = Boltzmann constant = 1.38 × 10^-23 Jmol^-1K^-1

T = absolute temperature

Using the equation (3), we can rewrite equation (1) as,

…(4)

Putting the values in equation (4) we get,

λ = 0.72 × 10^-10m

Using ideal gas equation we have,

PV = RT

PV = kNT

Where,

V = volume

N = number of moles

P = pressure

k = Boltzmann constant

T = absolute temperature

Mean distance between, r is given by,

r = 3.3 × 10^-9m

The mean separation between the atoms is greater than the De- Broglie wavelength.

Given:

Temperature, T = 27° C = 300 K

Mean separation between electrons, r = 2 × 10^-10 m

De-Broglie wavelength of electron is given by,

…(1)

Where,

h = Planck’s constant = 6.6 × 10^-34 Js

m = mass of electron = 9.1 × 10^-31 Kg

k = Boltzmann constant = 1.38 × 10^-23 Jmol^-1K^-1

T = absolute temperature

Putting the values in equation(1), we get,

λ = 6.2 × 10^-9 m

The De-Broglie wavelength is much larger than the interelectronic separation.

Quarks inside protons and neutrons are known to have fractional charges. Quarks do not show up in Millikan's oil drop experiment as these fractional charges do not exist independently, rather held together by strong nuclear force. These fractional charges are not found free in nature.

e/m or the specific charge is a important ratio as it is found in basic equations of electro-magnetism, for instance,

evB = mv²/r …(1)

→ (e/m)vB = v²/r

v = Br(e/m)

eV = 1/2mv² …(2)

→ v =

We see that the ratio (e/m) has an impact on the dynamics of electron in magnetic field hence e/m is given more importance that e or m.

At atmospheric/ ordinary pressures the Ions of gases move randomly and collide with each other , this random motion prevents it from setting a definite current in given direction, at low pressures these random motions decrease and if a potential difference is created between two points the gaseous Ions are able to move in definite direction setting a current. Hence, the gases become conductor at very low pressures.

Work function is the minimum energy required to set off the photo-electric emission. As the incident radiation falls on electrons they absorb defined wavelengths and if they gain enough energy they jump out of metal. But the energy absorption depends is very specific and depends on the initial energy level of electrons. Since all electrons are not in same energy level, all of them can’t absorb the monochromatic wavelength. Since different electrons start from different energy levels so there a energy distribution of photo-electrons.

We understand that momentum and wavelength of photons have physical significance as it is evident from X-diffraction crystallography. But Energy of photon which is related to the frequency is arbitrary to within an additive constant, hence absolute value of frequency has no physical significance.